# How do you find the derivative of  y = (2cosx)/(x+1)?

Mar 1, 2017

$= \frac{\left(- 2\right) \left(x \sin \left(x\right) + \sin \left(x\right) + \cos \left(x\right)\right)}{{x}^{2} + 2 x + 1}$

#### Explanation:

Recall that we use the quotient rule to differentiate a function of the form $\frac{P \left(x\right)}{Q \left(x\right)}$.

The quotient rule states that:
$\frac{d}{\mathrm{dx}} \frac{P \left(x\right)}{Q \left(x\right)} = \frac{Q \left(x\right) \cdot P ' \left(x\right) - P \left(x\right) \cdot Q ' \left(x\right)}{Q \left(x\right)} ^ 2$

I usually remember it as $\frac{B T ' - T B '}{B} ^ 2$ where $T$ is the top function and $B$ is the bottom function.

Now, applying the quotient rule to the function, we get:

$\frac{d}{\mathrm{dx}} \frac{2 \cos x}{x + 1}$$= \frac{\left(x + 1\right) \left(- 2 \sin x\right) - \left(2 \cos x\right) \left(1\right)}{x + 1} ^ 2$

$= \frac{- 2 x \sin \left(x\right) - 2 \sin \left(x\right) - 2 \cos \left(x\right)}{{x}^{2} + 2 x + 1}$

or
$= \frac{\left(- 2\right) \left(x \sin \left(x\right) + \sin \left(x\right) + \cos \left(x\right)\right)}{{x}^{2} + 2 x + 1}$